Find the mass of ammonia in the 100 cubic feet tank having a pressure of 70 psi at 120 °F.
Ideal gas law
PV=nRTV=100 ft3=2831.68 LP=70 psi=4.7632 atmT=120 °F=322.039PV=nRT \\ V=100 \;ft^3 = 2831.68 \;L\\ P=70 \;psi = 4.7632 \;atm\\ T= 120 \;°F = 322.039PV=nRTV=100ft3=2831.68LP=70psi=4.7632atmT=120°F=322.039
R=0.082058 L×atm/(K×mol)
n=PVRT=4.7632×2831.680.082058×322.039=510.40 moln = \frac{PV}{RT} \\ = \frac{4.7632 \times 2831.68}{0.082058 \times 322.039} \\ = 510.40 \;moln=RTPV=0.082058×322.0394.7632×2831.68=510.40mol
M(ammonia) = 17.031 g/mol
m(ammonia) =510.40×17.031=8692.62 g=8.692 kg= 510.40 \times 17.031 = 8692.62 \;g = 8.692 \;kg=510.40×17.031=8692.62g=8.692kg
Answer: 8.7 kg
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